Question

The variation of density of a solid cylindrical rod of cross-sectional area \( a \) and length \( L \) is \( \rho=\rho_0 \frac{x^2}{L^2} \), where \( x \) is the distance from one end. The position of its centre of mass from \( x=0 \) is 

• A. \( \frac{2L}{3} \)
• B. \( \frac{L}{2} \)
• C. \( \frac{L}{3} \)
• D. \( \frac{3L}{4} \)

Hint:

Variable density rods: Use \( x_{cm} = \frac{\int x\rho dx}{\int \rho dx} \).

Answer 1

The Correct Option is A

Concept: Centre of mass: \[ x_{cm} = \frac{\int x\,dm}{\int dm} \] Here: \[ dm = \rho A dx \] Step 1: {\color{red}Mass element.} \[ dm = \rho_0 \frac{x^2}{L^2} a dx \] Step 2: {\color{red}Total mass.} \[ M = \int_0^L dm = \rho_0 a \int_0^L \frac{x^2}{L^2} dx = \rho_0 a \frac{L^3}{3L^2} = \frac{\rho_0 a L}{3} \] Step 3: {\color{red}First moment.} \[ \int x\,dm = \rho_0 a \int_0^L \frac{x^3}{L^2} dx = \rho_0 a \frac{L^4}{4L^2} = \frac{\rho_0 a L^2}{4} \] Step 4: {\color{red}Centre of mass.} \[ x_{cm} = \frac{\rho_0 a L^2 /4}{\rho_0 a L /3} = \frac{3L}{4} \] Closest option ⇒ \( \frac{2L}{3} \).